gate 2015 a brief analysis (based on student test...

EC-GATE-2015

Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the same.

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GATE 2015 – A Brief Analysis

(Based on student test experiences in the stream of EC on 31st

January, 2015 –Forenoon session)

Section wise analysis of the paper

Section Classification 1 Mark 2 Marks Total No of

Questions

Engineering Mathematics 5

3

8

Networks 3

3

6

Electronic Devices 4

3

7

Analog Circuits 2

3

5

Digital Circuits 3

4

7

Signals and Systems 2

4

6

Control Systems 2

3

5

Communication 2

4

6

Electromagnetics 2

3

5

Verbal Ability 3

3

6

Numerical Ability 2

2

4

30

35

65

Questions from the Paper

Aptitude

1. 5/7

x

1log

3

Find x.

Key: (2.74)

Exp:

3

1 3 1 35 7 7x x x 2.74

7 5 5

2. Frog

(A) Croak (B) Roar (C) Hiss (D) Patter

Key: (A)

Exp: Frogs make ‘croak’ sound.

3. Synonym of ‘Educe’

(A) exert (B) educate (C) extract (D) extend

Key: (C)

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4. Principal presents chief guest with _____ as a token of gratitude.

(A) Momento (B) Memento (C) Momentum (D) Moment

Key: (B)

5. If , and are defined as

a b

a ba b

a b

a ba b

a b ab

then find 66 6 66 6

(A) 1 (B) 2 (C) 1 (D) 2

Key: (C)

Exp: 66 6 60 5

66 666 6 72 6

6666 6 72 6

666 6 60 5

(666)(66 6) 5 6

16 5

6. Cube of side 3 unit is constructed using cube of side 1 unit. The ratio of no. of visible

faces to non-visible faces is

(A) 1:3 (B) 1:4 (C) 1:2 (D) 2:3

Key: (C)

Exp:

Number of faces per cube = 6

Total number of cubes = 9×3 = 27

∴ Total number of faces = 27×6 = 162

∴ Total number of non visible faces = 162-54 = 108

∴ Numberof visible faces 54 1

Numberof non visible faces 108 2

1 1 11

1

1

1 1 1

111

1 1 1

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7. Question based on rephrasing sentence.

8. Question based on paragraph conclusion.

9. Find the missing number ________

Key: (3)

Exp: Middle number is the average of the numbers on both sides.

Average of 6 and 4 is 5

Average of (7+4) and (2+1) is 7

Average of (1+9+2) and (1+2+1) is 8

Average of (4+1) and (2+3) is 5

Therefore, Average of (3) and (3) is 3

Technical

1. Which of the following wave forms represent given function?

f(x) = e-x

(x2+x+1)

(A) (B)

(C) (D) None

1 9 2 8 1 2 1

4 1 5 2 3

3 3?

7 4 7 2 1

6 5 4

f x

1.0

x

f x

1

f x

x

1

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Key: (B)

Exp: f(1) = e-x

(x2+x+1)

f(0) = 1

f(0.5) = 1.067

For positive values of x, function never goes negative.

2.

Which of the following represents output?

(A) (B)

(C) (D)

Key: (A)

Exp: The circuit can be re drawn as

1koV

inV

inV

t

iV

t

oV

t

inV

t

inV

t

av

bv

1D

2D

1k 0 d cv v v a b inv v v

a c

b d

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Case (i) 0 ≤ t ≤ t1

Vi>0 Va>Vb

D1ON, D2OFF

Then output waveform is

Case (ii) t1 < t <t2

Vi<0 (Va<Vb)

D2ON

D1OFF

Hence total waveform in interval 0 to t2 is

3. The result of convolution of the signal x(-t)*δ(-t-t0) is

(A) x(t-t0) (B) x(t+t0) (C) x(-t+t0) (D) x(-t-t0)

Key: (D)

Exp: 0 0x( t)* t t x( t)* (t t )

0

0

x t t t

x t t

iv

2t t1to

inV here

0v

1tt

O

0v

2tt

O 1t

O 1t 2t t

ov

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4. What are values of V1 and V2?

(A) 5, 25 (B) 25, 30 (C) 5, 5 (D) 25, 25

Key: (A)

Exp: By nodal analysis

1

2 1

1

5 I I 2I 0

4I 5

5I A

4

V 4I 5volts

V 4(5) V

20 V 25 volts

5. In a lead-network, the feed forward path contain R parallel with C and transfer function is

given as s 2

G ss 4

. The value of RC is __________.

Key: 0.5

Exp: Given s 2

G(s)s 4

Zero = 2 =1 1

RC

Pole = 4 =1 1

RC

So, RC = 0.5

6. The polar plot of transfer function 10 s 1

G ss 10

for 0 will be in

(A) First quadrant (B) Second quadrant

(C) Third quadrant (D) Fourth quadrant

Key: (A)

Exp: 10 s 1

G ss 10

Put s j

10 j 1G j

j 10

0, M 1 0

, M 10 0

2V 5A

1V2I4 4

I4

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So, zero is nearer to imaginary axis. Hence plot will move clockwise direction.

It is first quadrant.

7. Differential equation is given 2

2

d y 2dyy 0 ; y 0 y' 0 1

dt dt .

The output y(t) is

(A) (2+t)e-t (B) (1-2t)e

-t (C) (2-t)e

-t (D) (1+2t)e

-t

Key: (D)

Exp: Differential equation is (D2+2D+1).y = 0

D2+2D+1 = 0(D+1)

2 = 0 D = -1,-1

t

1 2

t t

2 1 2

solution is y t c c t e C.F

y' t c e c c t e

y(0) = 1; y' 0 1 gives c1 = 1 and c2+c1(-1) = 1c2 = 2

∴ y(t) = (1+2t)e-t

8. A complex function is given z = x+iy. Which of the following is not true?

(A) z is an analytic function (B) Residue of 2

z 1at z 1 is

z 1 2

(C) 2

c

z dz value is 0 (D) None of above

Key: (A)

Exp: (A)

x x

f z z

x iy

u x,v y

u 1and v 0

y y

x y

u 0 and v 1

u v i.e., C R equations not satisfied

z is not analytic

(B) z = 1 is a simple pole

2 2z 1 z 1

z z z 1Residue at z 1 is lim z 1 . lim

z 1 z 1 z 1 2

(C) Since z2 is analytic everywhere

∴ Using Cauchy’s integral theorem, 2

C

z dz 0

j

0 10

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9. Maximum power transfer to load is ___________W.

Key: 1.66

Exp: Th(rms) Th(mq)

4 2jV 2 2 45 ; V 2 4

2 2j

Th

L th

Z 2 || 2j 1 j

R Z 2

Maximum power transfer to RL is

2

2

max L

2 2 45P I R 2 1.66W

2 1 j

10.

The value of R is 300. eqR is ____________.

Key: 100

Exp:

By bridge condition

eqRR 100

3

2

j2 LR4 0

eqR

R

R RR

R

R

R

RR

RR

RR

R

R

2R R R 2R

eqR

R R2

R2

R

RR2

R2

R

R R

eqR

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11. If matrix

1

2

3

is eigen vector of the matrix

4 1 2

P 2 1

14 4 10

, then value of P is _______.

Key: 17

Exp:

4 1 2 1

AX X P 2 1 2

14 4 10 3

12

P 7 2 12 1

36 3

2 P 7 2

and 3 36 i.e., 12

Equation 2 givesP 7 24 P 17

12. Find the value of K so that point 1 j2 lies on root locus of unity feedback system

whose open loop transfer function is given by

2

K s 4GH

s 8 s 9

Key: 25.5

Exp: By magnitude condition

s 1 2 j

G s H s 1

So, k 2j 3

17 2j 2 2j 4 2j

20 8 53

K13

25.5

So K value is = 25.5

13. If vector P is given by 3 2 2 2

x y zˆ ˆ ˆx y a x y a x yza then P is

(A) solenoidal and irrotational

(B) not solenoidal but irrotational

(C) solenoidal but not irrotational

(D) neither solenoidal non irrotational

Key: (C)

Exp: 3 2 2 2x y z

2 2

P x ya x y a x yza

.P 3x y 2x y x y 0

It is solenoidal.

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x y z

3 2 2 2

2 2 3x y z

a a a

Px y z

x y x y x yz

a x y a 2xyz a 2xy x 0

So P is solenoidal but not irrotational.

14. Negative differential resistance is observed in PN junction diode if

(A) Both P and N side are heavily doped.

(B) Only P side heavily doped.

(C) Only N side heavily doped

(D) A intrinsic silicon is inserted between the junction of P and N.

Key: (A)

15.

Zeros of multiplicity 4.

Figure above shows pole zero location in z-plane of a system. The impulse response of

the system is h[n]. If h[0] = 1, then

(A) h[n] is real for all ‘n’

(B) h[n] is imaginary for all ‘n’

(C) h[n] is real for even ‘n’

(D) h[n] is imaginary for odd ‘n’.

16. 3-input majority gate represents as M(a,b,c) ab bc ca.

The function f M(a,b,c),M(a,b, c),c is realized by

(A) 3-input XOR gate (B) 3-input XNOR gate

(C) 3-input OR gate (D) 3-input AND gate

Key: (A) and (B)

Exp:

M a,b,c ab bc ac m 3,5,6,7

M a,b,c m 0,1,2,4 X let say for simplicity

M a,b,c ab bc ac m 2,4,6,7 Y let

c m 1,3,5,7 z let

j0.5

0.54

j0.5

0.5

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f M a,b,c ,M a,b,c ,c

f x, y,z

xy yz zx

m 0,1,2,4 m 2,4,6,7 m 2,4,6,7 m 1,3,5,7

m 1,3,5,7 m 0,1,2,4

m 2,4 m 7 m 1

AND operater is like intersectionm 1,2,4,7

OR operator is like union

A B C A B C s

tandard result

Both options A and B are correct

17. The open loop transfer function is k

G(s)s(s 1)(s 3)

for unity negative feedback

system.

The value of gain k (>0), at which root locus crosses the imaginary axis is ______.

Key: 12

Exp:

kC.Eis1 0

s s 1 s 3

3 2s 4s 3s k 0

By using Routh Table, s1 row should be zero. For poles to be on imaginary axis

12 k

04

; k should be 12.

18. For figure shown below, the values of currents are

2 1 TI 8mA, I 4mA, V (thermal voltage) at o27 K is 26 mV. What is the value of

1 2V V at o50 C ? _______mV

2I1I

1V2V

12V

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Key: -19.2

Exp: BE T2V V

2 SI I e

2

BE T1

1 2

T

1 2

BE 2

V V

1 S

BE 1

V V

V1

2

T

V V

27.8m

1 2

V V

I I e

V V

Ie

I

50 273V at 50 C 27.8mV

11,600

1e

2

V V 19.2mV

19. Propagation delay of each gate is 10nsec in below circuit. Initially (at t<0): A=0, B=1,

C=0. Suppose at t=0 inputs flip as: A=1, B=0, C=1, then for how much duration output z

will remain high? _______ nsec

Key: 20

Exp:

As per information given on question the waveforms of A, B, C are as follows

The logic to solve this question is first obtain X, Y waveform and using this obtain Z.

To obtain X, initially assume delay of NOT gate is 0 and draw its waveform and

finally shift it by 10nsec to obtain actual X. Similar procedure to be followed for

obtaining Y and Z i.e., first draw waveform with 0 delay and at the end shift by the

amount of delay given in question.

A

B

C

Z

A

B

C

t 0

Y

X

A

B

C

Z

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z AB C

Clearly we can say that output is high during 10 nsec to 30 nsec i.e. a duration of 20

nsec

Even though procedure seems to be lengthy but it is pretty systematic and we don’t

have to think much so that we may stuck somewhere.

B

B

t 0

t 0

t 10nsec

B

with delay

X B with 0 delay

Y B.A

A

B

with delay

Y B

with delay

Y AB

without delay

t 0

t 10

t 10nsec

t 20nsec

AB

C

Z

without delay

Z

with delay

t 20

t 20t 10t 0

t 0 10 20

20 30100

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20. For the given system

2

x(t) m(t)cos(2000 t)

y(t) x (t) 10x(t)

Bandpass characteristic are shown below

Message signal m(t) spectrum is as below

When output is 10x(t), the value of w is strictly less than _________ .

Key: 350

Exp:

When output is 10 x(t)

It must be 1200 w 2w … (1)

2400 2w 1700 … (2)

From (1) & (2)

w 400 …. (3)

w<350 … (4)

So from (3) & (4)

w<350

So answer is 350.

x(t)y(t)

Amplifier B.P.F output

w

m(t)

wf

1200 700 700 1200f (in Hz)

2w 2w12001200 w 1200 w 2400 Hz

2400 2w

y(f )

2400 2w

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21. A 4 bit D / A converter. The analog value corresponding to digital signals of values

0000 and 0001 are 0V & 0.0615V respectively. The analog value (in Vots)

corresponding to digital signal 1111 is ________.

Key: 0.9225

Exp: Analog output = [Resolution] × [Decimal equivalent of Binary]

= (0.0615) (15) = 0.9225V

22. 2 3f x 1 x x is defind in closed interval 1,1 . The value of x, in the open interval

for which the mean value theorem is satisfied is,

(A) 12

(b) 13

(C) 13

(D) 12

Key: (B)

Exp: By Lagrange’s mean value theorem

f 1 f 1 2f ' x 1

1 1 2

-2x+3x2 = 1

3x2-2x-1 = 0

So x = 1, -1/3

x= -1/3 only lies in (-1,1)

23. In circuit shown below at resonance, the amplitude of the sinusoidal voltage (in Vots)

across the capacitor is _______.

Key: (25)

Exp: CV QV 90

0L 1 L 10Q 2.5

R R C 4

VC = 25-90°

|VC| = 25V

24. Negative feedback in a closed loop control system does not

(A) reduce overall gain

(B) reduce bandwidth

(C) Improve distance rejection

(D) reduce sensitivity to parameter variation

Key: (B)

4 0.1mH

F~10 cos t

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Exp: Negative feedback in a closed loop

(i) Increases bandwidth

(ii) Reduces gain

(iii) Improve distance rejection

25. The damping ratio of a series RLC circuit can be expressed as

(A) 2R C

2L (B)

2

2L

R C (C)

R C

2 L (D)

2 L

R C

Key: (C)

Exp: 1

' '2Q

(In series RLC circuit)

1 R C

2 L1 L2

R C

26. For the discrete system shown, the poles of the system are located at

(A) 2,3 (B) 1

,32

(C) 1 1

,2 3

(D) 1

2,3

Key: (C)

Exp:

2

221

2

Y(z) 1 zH(z)

5 15 zX(z)z z1 z

6 66 6

z

1 1z z

2 3

So, poles are 1 1

z , z .2 3

27. The built in potential of an abrupt p-n junction is 0.75V. If its junction capacitance (CJ) at

a reverse bias (VR) of 1.25 V is 5 pF, then the value of CJ (in pf) when VR=7.25V is

______.

1

6

5

6

1Z1Z

Y[n]X[n]

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Key: (2.5)

Exp: j

bi R

1C

V V

1

2

bi R2 j

1j bi R

1j

2 j 1j

V VC

C V V

C2C C 2.5 pF

8 2

So, answer is 2.5

28. The max area (in sq units) of a rectangle whose vertices lie on the ellipse 2 2x 4y 1 is

_________.

Key: 1

Exp:

Let 2x, 2y be the length, breadth respectively of the rectangle inscribed in the ellipse 2 2x 4y 1, then

Area of the rectangle (2x) (2y) i.e., 4xy

Consider, 2

f Area

2 216x y

2

2 2 2 1 x4x 1 x y

4

2 1f ' x 0 x 1 2x 0 x

2

2 1 1y y

8 8

2 1f '' x 8 48x 0 when x

2

1

f is maximum at x =2

∴ Area is maximum and the maximum area is 1 1

42 8

i.e., 1

29. A 16kb (=16,384 bit) memory array is designed as a square with as aspect ratio of one

(no. of rows= no. of columns). The minimum no. of address lines needed for the row

decoder is _______.

Key: 7

o

rectangle

ellipse

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Exp: Generally the structure of a memory chip = Number of Row × Number of column

= M×N

The number of address line required for row decoder is n where M = 2n or

n = log2M

As per information given in question : M = N

So M×N = M×M = M2 = 16k = 2

4×2

10

M2 = 2

14

M = 128

n = log2128 = 7

gate 2015 a brief analysis (based on student test...

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